Problem: The value of $\sqrt{6}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Answer: Consider the perfect squares near $6$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $4$ is the nearest perfect square less than $6$ $9$ is the nearest perfect square more than $6$ So, we know $4 < 6 < 9$ So, $\sqrt{4} < \sqrt{6} < \sqrt{9}$ So $\sqrt{6}$ is between $2$ and $3$.